Lie-algebra representation powers - plethysms

[lpetrich]

As I've posted earlier, I've succeeded in finding expressions for the projection matrices for subalgebras in all but the most general case of SU/SO/Sp(n) -> simple algebra with n-D rep

To illustrate what the problem in that case is, let's work out some simple examples:

First, SU(6) -> SU(3)
Antisymmetric rep powers (plethysms):
1: 10000 -> 20
2: 01000 -> 21
3: 00100 -> 30 + 03
4: 00010 -> 12
5: 00001 -> 02
6: singlets, 7+: none
Rep products:
10000 * 00001 -> 10001 + 00000
20 * 02 -> 22 + 11 + 00
Adjoint + singlet in both cases

Powers 1,2,4,5 suggest unambiguous mappings from the original weight space to the subalgebra one. Power 3 has an ambiguity. How to resolve it?

Trying SO(7) -> G2 gives plethysms
1: 100 -> 01
2: 010 -> 10 + 01
3: 002 -> 02 + 01 + 00
4 ~ 3, 5 ~ 2, 6 ~ 1, 7: singlets, 8+: none
When one gets to the original algebra's spinor rep, one gets a 2 instead of 1. Fortunately, there's a rep with 2 for the subalgebra.

Trying SO(8) -> SO(7) gives plethysms
1: 1000 -> 001
2: 0100 -> 100 + 010
3: 0011 -> 101 + 001
4: 0020 + 0002 -> 200 + 100 + 002 + 000
5 ~ 3, 6 ~ 2, 7 ~ 1, 8: singlets, 9+: none
Even worse. In addition to the subalgebra ambiguities, there's a spinor-rep ambiguity on the original-algebra side.

 

[lpetrich]

I think that I solved the reality problem for subalgebra reps. To do that, I have to step back a bit and be more general. Algebra generators L are related by

L(subalgebra)_i = sum_j c_ij L(original)_j

where we can choose L to be Hermitian. This makes the c's real. Take the conjugate. This makes:
conjugate of original rep -> conjugate of subalgebra rep

A rep is self-conjugate if L^* = Z.L.Z^-1 for all L defined in the rep. The rep is real if
Z.Z^* = I
and pseudoreal if
Z.Z^* = - I
Must be one or the other if irreducible; can also be mixed if reducible.

Thus,
real -> real
pseudoreal -> pseudoreal
complex -> any reality: real, pseudoreal, or complex

The adjoint rep can be constructed with a basis space being the algebra generators themselves: |L>. Original -> subalgebra:
|L(subalgebra)_i> = sum_j c_ij |L(original)_j>

Thus,
adjoint rep -> rep that contains the adjoint rep

-

For finite groups, it's possible to prove that every irrep and its conjugate has one copy of the identity rep in its product. The proof. Let
n(r1,r2,r3) = (1/N)*sum_a char(r1,a) * char(r2,a) * char(r3,a)
over all N elements a for irreps r1, r2, r3.

The product of r1 and r2 contains n(r1,r2,r3*) copies of r3.

Let r2 = r1*, its conjugate and let r3 = identity rep. Then, the sum for n reduces to
(1/N)*sum_a |char(r1,a)|^2 = 1 by orthonormality.

For compact Lie groups, one replaces the sum over elements a by an integral over the Haar measure of the element parameters, and N by that integral over 1. Orthonormality carries over, and that result appears here also. Since it is true for the group, it must also be true for the algebra.

Reality: a rep D(a) is self-conjugate if there's some matrix T such that
T_ijD_ik(a)D_jl(a) = T_kl
It's a real rep if T is symmetric, psuedoreal if T is antisymmetric. It carries over into Lie algebras:
T_kjL_ki + T_ikL_kj = T_ij
The T is essentially a singlet rep of the algebra.

So all these products contain exactly one singlet rep:

• Real: symmetric square
• Pseudoreal: antisymmetric square
• Complex: product with conjugate

 

[lpetrich]

I've been stumped on getting the adjoint reps from powers of reps and products of reps with conjugates, but it's possible to get them in some special cases:

• SU(n): product of n-D rep with its conjugate minus a singlet
• SO(n): antisymmetric square of n-D rep: antisymmetric 2-tensor
• Sp(n): symmetric square of n-D rep: symmetric 2-tensor
• Exceptional fundamental reps: G2, F4, E8: in antisymmetric square, E7: in symmetric square, E6: in product with conjugate
• Adjoint rep of every algebra: in antisymmetric square

For SO(n) spinors, one can construct an antisymmetric p-tensor with some spinors:
(spinor).(antisymmetric product of p Dirac matrices).(spinor)

There are n SO(n) Dirac matrices, and one gets the adjoint for p = 2.The plethysm method of calculating subalgebra projection matrices not only has some ambiguities, it's computationally expensive with my brute-force code. For a rep with size n and going up to power p requires O(n^p/p!) calculations. Its maximum is at p ~ n, giving O(eⁿ/sqrt(2*pi*n)) calculations.

I've worked out how to do plethysms on SU(n) reps using Young-diagram techniques, but it's rather gruesomely complicated, and I don't know how to generalize it to SO(n), Sp(n), or the exceptional algebras. Any Young-diagram techniques for doing SO(n) products? One could work with their SU(n) supergroups, but I don't know how well that would be justified.

For instance, a symmetric traceless 2-tensor in SO(n) is (2,0,0,...), and it's SU(n) (2,0,0,...) - (0,0,0,...), or in Young diagrams, (2) - (). One would use Young-diagram techniques, then go back from SU(n) to SO(n).In general, rep addition and multiplication form a commutative semiring, as the nonnegative integers do.

Its elements are every possible rep of a group or algebra. Each one can be decomposed into a set of nonnegative-integer number of copies of each irrep, and that can indeed be interpreted as a vector of nonnegative integers with each irrep being a basis vector.

Its addition and multiplication are rep sums and products. It has an additive identity, the empty rep, and a multiplicative identity, the singlet or scalar rep.

 

[lpetrich]

I think that I now understand Weyl orbits, and I've moved them into the main code of the Mathematica version. I'm now working on Weyl orbits for the Python and C++ versions.

To understand them, let's first consider the Weyl groups. This is the group of symmetries of an algebra's root vectors. An algebra's group is generated by operators T for root a operating on vector x:
T(a).x = x - 2*(x.a)*a/(a.a)

In the Cartan-Weyl basis, what my code primarily uses, the structures of the Weyl groups are not very clear. But one can use alternate bases, ones where the algebra metric becomes the identity matrix. Here, the a(i)'s will be the simple roots and the e(i)'s the basis vectors.

Cartan-Weyl: a(i) = e(i). The metric is only the identity matrix in the case of rank 1.

Identity-matrix versions:
A(n): a(i) = e(i) - e(i+1) -- (n+1) basis vectors
B(n): for i = 1 to n-1: a(i) = e(i) - e(i+1); a(n) = e(n)
C(n): for i = 1 to n-1: a(i) = e(i) - e(i+1); a(n) = 2*e(n)
D(n): for i = 1 to n-1: a(i) = e(i) - e(i+1); a(n) = e(n-1) + e(n)

The exceptional algebras are more complicated.
G2: a(long) = 2*e1 - e2 - e3; a(short) = -e1 + e2
or a(long) = -3*e1 + sqrt(3)*e2; a(short) = 2*e1
F4: a1 = e1 - e2; a2 = e2 - e3; a3 = e3; a4 = - (1/2)*(e1 + e2 + e3 + e4)
E6, E7, and E8 are even worse.

Now the Weyl groups in those bases.

A(n): for a(i): interchange coordinates i and i+1
The group: permutations of identity matrix with size (n+1), thus Sym(n+1)
Order: (n+1)!
The symmetry group of the n-D regular simplex (generalized triangle and tetrahedron).

B(n) and C(n): the A(n-1) generators with a(n) reversing the sign of coordinate n.
The group: the A(n-1) group multiplied by the group of diagonal matrices with n 1's and -1's.
Order: 2ⁿ * n!
The symmetry group of the n-D hypercube (generalized square and cube) and cross polytope (generalized square and octahedron).

D(n): the A(n-1) generators with a(n) interchanging coordinates n-1 and n and also reversing their sign.
Like for B(n) and C(n), but with an even number of -1's.
Order: 2^n-1 * n!
Consider a hypercube's vertices in a coordinate system where their positions are vectors of +-1's. The D(n) symmetry group is the symmetry group of those with even numbers of -1's, and also those with odd numbers of -1's.

G2: Dih(6), the symmetry group of the regular hexagon.
Order: 12

F4: The group for B(4) with 4*4 nonsingular matrices of +- 1/2 added.
Order: 1152
The symmetry group of the 4D 24-cell.

E6, E7, and E8 have more complicated groups, with orders 51840, 2903040, and 696729600, respectively.

 

[lpetrich]

Now for what a Weyl orbit is. It's a group-theory kind of orbit. Consider a group action: for some g in group G, find x' = g(x) for some x and x' in some set X. Thus: X = G(X).

If the group is a matrix group, then X can be a set of vectors, with action x' = g.x -- the dot or inner product.

If one starts with some x, then for all group elements g, the g(x)'s form an "orbit". The order or size of an orbit is related to that of the group by the orbit-stabilizer theorem:

(order of group) = (order of orbit) * (order of "stabilizer" subgroup, the subgroup that does x = g(x)).

So orbits can be much smaller than groups, and that's what one often finds here.


For roots in algebras and their reps, the orbits generated by the Weyl groups are Weyl orbits. The roots in the orbits can be related to weights, and the weights have the property that only one of them has all-nonnegative components: the dominant weight. So as one designates irreps by their highest weights, one can do likewise with Weyl orbits.

One can find the Weyl orbits in irreps using the same procedure as for rep roots. All one has to do is keep those with nonnegative weight vectors, and stop when the solution process fails to find any more. Multiplicities / degeneracies one can find with Frobenius's theorem, working from the dominant weights' roots. However, some of the roots that one uses in it will be non-dominant, and one has to find which orbits those roots are in. But that does not present much difficulty, and finding an irrep's orbits is usually much faster than finding its basis set of roots/weights.

This speedup is not original with me -- I read about it in [1206.6379] LieART -- A Mathematica Application for Lie Algebras and Representation Theory. Though that package has much fancier output options than mine, it does not seem to do plethysms.

LieART also mentioned another speedup for doing rep products and subalgebra reps. One still needs to use complete expansions of the input reps, but as one calculates the products and subalgebra reps, one can keep only the roots with nonnegative weights, because those will designate the Weyl orbits. One can then find what irreps by using their Weyl-orbit content.


Now for expanding a Weyl orbit. One does not need an entire Weyl group for that, just its generators, and my code uses only the generators for the algebra's simple roots. One can start with the dominant root/weight and work one's way downward until one can proceed no further. Likewise, to see what orbit a root is in, one can work one's way upward until one gets to the dominant root/weight.


Using those alternative basis vectors for the roots of A(n), B(n), C(n), and D(n), one can use Weyl-group elements implicitly to get further speedups when expanding Weyl orbits.

• Take the dominant root from the Cartan-Weyl basis into an appropriate alternative basis.
• Apply the Weyl-group elements implicitly by doing sign changes and permutations, generating all the roots in the orbit.
• Take these roots back into the Cartan-Weyl basis.

However, that does not work so well for the exceptional algebras. One can easily special-case G2, but F4 is more difficult. I've been unable to find anything even halfway simple for E6, E7, or E8.

At least in my Mathematica version, I've found speedups for A(n), B(n), C(n), and D(n), but not for G2 or F4.

 

[lpetrich]

Back to the question of real vs. pseudoreal vs. complex and what reps are present, I've come across this interesting result.

(rep) * (conjugate rep) contains both the scalar and the adjoint.
If the rep is real, then the symmetric part contains the scalar and the antisymmetric part the adjoint.
If the rep is pseudoreal, then the symmetric part contains the adjoint and the antisymmetric part the scalar.
There is one each in every case.

Here's a proof:
Start out with a Lie-algebra transformation: V -> (1 + i*ε*L).V = V + i*ε*D(V) Thus,
D(V) = L.V
D(V^*) = - L^*.V^* -- conjugate rep
L is Hermitian: L^* = L^T

Now form a rep from the outer product of V and V^*: T, with indices (V index, V^* index)

D(T) = L.T - T.L
For T = scalar I, D(I) = 0
For T = another algebra generator L', D(L') = [L,L'] -- one gets the adjoint rep.

-

Now the self-conjugate case.

V^* = Z.V where Z is some constant Hermitian matrix.
L^* = L^T = - Z.L.Z^-1

For the square rep for V, T has indices (V index, V index), and
D(T) = L.T + T.L^T = L.T - T.Z.L.Z^-1
D(Z^-1) = 0 -- scalar, since Z is a constant.
D(L'.Z^-1) = [L,L'].Z^-1 -- adjoint, since L' is an algebra generator.

So we now need to consider the symmetry properties of Z^-1 and L.Z^-1 -- scalar and adjoint reps.

To do that, we will need the symmetry properties of Z. We start with
L^* = - Z.L.Z^-1

Complex conjugate again:
L = (Z^*.Z).L.(Z^*.Z)^-1
giving Z^*.Z = Z^T.Z = s*I where s > 0 for real and s < 0 for pseudoreal.

Hermitian conjugate this time:
L^* = - Z^-1.L.Z
L = (Z^*.Z^-1).L.(Z^*.Z^-1)^-1
Z^*.Z^-1 = t*I
Z^* = Z^T = t*Z
Z = t*Z^T
Thus, t = +- 1 and Z² = s*t
Since Z is Hermitian, its eigenvalues are real and those of Z² are greater than 0, thus, s has the same sign as t.

-

Let's consider the symmetry of L.Z^-1. Taking the transpose gives Z^T-1.L^T = - (Z^T-1.Z).(L.Z^-1) = - t*(L.Z^-1)

Likewise, the transpose of Z^-1 is t*Z^-1.

For real irreps, t = 1 and s > 0, and the symmetric product gives the scalar and the antisymmetric one the adjoint.

For pseudoreal irreps, t = -1 and s < 0, and the symmetric product gives the adjoint and the antisymmetric one the scalar.